Medal + Fan to the person who comes up with the best explanation for d/dx lnx = 1/x!!!!!!!!!!!!
I can't fan KA, though!!!!!!!!!!!!!!!!!!!!!!
The derivative of ln x=1/x, that's the formula, didn't you know that?
Do you know it? Or do you just believe what people tell you?
That's the fact.
Prove it then
Sorry, I don't have time for that and I'm not strong in proofs.
That's like the evolution/creation debate. People believe that we came from a rock 4.6 billion years ago only because people tell them it. Makes no sense!
Who knows what exactly happened 4.6 billion years ago?
So prove that d/dx lnx = 1/x
$$e^{\log x}=x\\\frac{d}{dx}e^{\log x}=\frac{d}{dx}x\\e^{\log x}\frac{d}{dx}\log x=1\\\frac{d}{dx}\log x=\frac1{e^{\log x}}=\frac1x$$
This follows from the proof I gave you for \(\frac{d}{dx}e^x=e^x\) last time :-p
note that to get the third line I applied the chain rule to the left-hand side
\[f(x)=\ln x\\ \begin{align*}\frac{d}{dx}f(x)&=\lim_{h\to0}\frac{\ln(x+h)-\ln x}{h}\\ &=\lim_{h\to0}\frac{\ln\left(\dfrac{x+h}{x}\right)}{h}\\ &=\lim_{h\to0}\frac{\ln\left(1+\dfrac{h}{x}\right)}{h} \end{align*}\] Substitute \(t=\frac{h}{x}\) so that \(t\to0\) as \(h\to0\): \[\begin{align*}\frac{d}{dx}f(x)&=\lim_{t\to0}\frac{\ln\left(1+t\right)}{xt}\\ &=\frac{1}{x}\lim_{t\to0}\frac{\ln\left(1+t\right)}{t} \end{align*}\] Next, you can use the fact that \(\ln(1+t)\approx t\) for \(t\) near 0: \[\begin{align*}\frac{d}{dx}f(x) &=\frac{1}{x}\lim_{t\to0}\frac{t}{t}\\ &=\frac{1}{x} \end{align*}\]
@SithsAndGiggles that's a better approach tbh :-p
Gotta use the definition when you can :)
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